3 Answers
3

I assume that you are talking about a faithful action of $\mathbb{Z}/n\mathbb{Z}$ on $\mathbb{Z}/p\mathbb{Z}$. First, there are the obvious lifts of the irreducible representations of $\mathbb{Z}/n\mathbb{Z}$ to $G$. For the rest:

Exercise 1: The induction of a non-trivial character of $\mathbb{Z}/p\mathbb{Z}$ to $G$ is irreducible.

Exercise 2: When are two such inductions isomorphic?

Exercise 3: Now count the sums of squares of the degrees of the characters that you get this way.

I have to say that this is rather tough homework if you were not given any of these hints. But with them, you should be able to do the rest yourself.

Edit: More generally, suppose that $G=A\rtimes H$ where $A$ is abelian. Then all irreducible characters of $G$ are obtained as follows. $H$ acts on the irreducible characters of $A$ by $(h\cdot\chi)(a) = \chi(h^{-1}ah)$. Let $\chi$ be a linear character of $A$. Extend it to $S_\chi=A\rtimes \text{Stab}_H(\chi)$, where $\text{Stab}_H(\chi)\leq H$ is the stabiliser of $\chi$ in $H$ under the above action, by setting $\chi(as) = \chi(a)$ for $a\in A,s\in \text{Stab}_H(\chi)$. Let $\rho$ be an irreducible character of $\text{Stab}_H(\chi)$, lift it to an irreducible character of $S_\chi$. Then $\text{Ind}_{G/S_\chi}(\chi\otimes \rho)$ is an irreducible character of $G$ and they all arise in this way. I will leave it to you to determine when two such inductions are isomorphic, so as not to spoil the homework exercise.

Note that your question is a special case of this, since $H=\mathbb{Z}/n\mathbb{Z}$ acts faithfully on $\mathbb{Z}/p\mathbb{Z}$, and therefore also on its irreducible characters. Thus, $\text{Stab}_H(\chi)$ is trivial in your case, whenever $\chi$ is non-trivial.

Hi Alex wahat is the degree of these irreducible characters?
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Sven WirsingJan 25 at 10:46

@SvenWirsing: In the specific situation of the OP, they are 1- and $n$-dimensional. In the general situation of my Edit, they are $($dim $\rho\cdot \#$orbit$_H(\chi))$-dimensional. That just follows from the fact that under induction, the dimension of a representation grows by the index of the subgroup that you are inducing from.
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Alex B.Jan 25 at 14:17

I investigate the sum of degrees of the irreducible characters. In the example here the derived group is $A$ and hence there are $n$ linear characters. Using the fact that $np$ is the sum of squares of the irr. characters I see that there are $\frac{p-1}{n}$ irr. of degree n. Hence the sum is $p-1+n$\\ Using your construction above one get from the trivial character of A and the n linear characters from H exactly n irr. characters of degree 1. From the other $p-1$ irreducibles of $A$ we get $p-1$ induced irr. of degree $n$ as the stabilizer is trivial in this cases. Where is my fault?
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Sven WirsingJan 25 at 19:33

You need to do what I labelled as Exercise 2.
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Alex B.Jan 26 at 10:32

Have a look at Weintraub - "Representation theory of finite group" in the section Mackey Machine. They give the right tool for exactly studying the irreducible representation of a semidirect product $H \rtimes G$, where $H$ is abelian.

Here is a place to start, although not a complete solution by any means. Let $G=\mathbb{Z}/p\mathbb{Z}\rtimes \mathbb{Z}/n \mathbb{Z}$. Then $\mathbb{Z}/n\subset G$ and $\mathbb{Z}\subset G$.

If we restrict a representation to one of these two subgroups, it will split as a direct sum of irreducibles, and because the subgroups are abelian, this means that we can find a basis of eigenvectors for the action of $\mathbb{Z}/p$ (or a basis of eigenvectors for the action of $\mathbb{Z}/n$).

Suppose that $\omega$ is a $p$th root of unity, and that $v$ is a vector such that $k.v=\omega^k v$ for $k\in \mathbb{Z}/p$. If $j\in \mathbb{Z}/n$, then how does $k$ act on $j.v$? Use the commutation relations that you know you have in $G$ between elements of $\mathbb{Z}/n$ and $\mathbb{Z}/p$